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Question Number 82018 by TawaTawa last updated on 17/Feb/20

$$\mathrm{Differentiate}\mathrm{y}=2^{\mathrm{x}}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{principle}.$$

Commented by john santu last updated on 17/Feb/20

$$lny=xln2$$$$\underset{h\to 0}{\lim }ln\left(y\right)=\underset{h\to 0}{\lim }ln\left(2\right)\frac{(x+h)-x}{h}$$$$\frac{y^{\prime }}{y}=ln\left(2\right)\underset{h\to 0}{\lim }\frac{h}{h}$$$$y^{\prime }=y.ln\left(2\right)=2^x.ln\left(2\right)$$

Commented by TawaTawa last updated on 17/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by TANMAY PANACEA last updated on 17/Feb/20

$$y+△y=2^{x+△x}$$$$△y=2^{x+△x}-2^x=2^x(2^{△x}-1)$$$$\frac{dy}{dx}=\underset{△x\to 0}{\lim }\frac{△y}{△x}=$$$$=\underset{△x\to 0}{\lim }\frac{2^x(2^{△x}-1)}{△x}$$$$=2^x\times \left(\underset{△x\to 0}{\lim }\frac{e^{△x.ln2}-1}{△x.ln2}\right)\times ln2$$$$=2^x\times 1\times ln2=2^{x}ln2$$

Commented by TawaTawa last updated on 17/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 17/Feb/20

$$y^{\prime }=\underset{h\to 0}{\lim }\frac{2^{x+h}-2^x}{h}$$$$=2^x\underset{h\to 0}{\lim }\frac{2^h-1}{h}$$$$=2^x\underset{h\to 0}{\lim }\frac{e^{h\ln 2}-1}{h}$$$$=2^x\underset{h\to 0}{\lim }\frac{[1+h\ln 2+\frac{{\left(h\ln 2\right)}^2}{2!}+\frac{{\left(h\ln 2\right)}^3}{3!}+...]-1}{h}$$$$=2^x\underset{h\to 0}{\lim }[\ln 2+\frac{h{\left(\ln 2\right)}^2}{2!}+\frac{h^2{\left(\ln 2\right)}^3}{3!}+...]$$$$=2^x\ln 2$$

Commented by TawaTawa last updated on 17/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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